How many atoms are in cacl2

Mass, amount of substance, atomic mass, molar mass

 

Dimensions

When a diamond burns completely with the mass M. = 1 gram (1g) in pure oxygen you can weigh the diamond, the required oxygen and also the reaction products. The physical base quantity mass is given in the unit kilogram or gram. From 1889 to 2019, the original kilogram stored in Paris was used as the reference weight. From May 20, 2019, the base unit kilogram will be derived from Planck's constant.


Physical base sizeDimensions
dimension
M.
Formula symbolm
unitkg, g



Amount of substance

In chemistry, substances are viewed as dynamic systems. When burning a diamond, you not only want to know how much oxygen is being used, but also the ratio of the atoms involved. This requires the physical base quantity of the amount of substance with the unit mole:


Physical base sizeAmount of substance
dimension
N
Formula symboln
unitmol



The definition, valid from May 20, 2019, specifies an absolute number: “One mole is 6.02214076 × 1023 Particles (atoms, molecules, ions, etc.). "This number is fixed, it is called Avogadro's constant NA..

The old definition, which was valid until May 19, 2019, derived the mole as a relative number depending on the carbon: The amount of substance 1 mole contains as many particles as there are atoms in 12 grams of isotopically pure C-12 carbon.


Atomic mass

The atomic masses of hydrogen or carbon atoms can now be determined relatively precisely through experimental arrangements. One twelfth of the mass of the carbon isotope C-12 is used as a reference value:

 

Reference value atomic mass

Formula symbolm
unitg, u
Reference value1u = 1.660539040 × 10−24 G
1u = one twelfth the mass of a C-12 atom


Compared to 1u, all atoms are heavier in their mass by approximately an integral multiple. This is because the values ​​of the neutron mass and the proton mass are each close to 1u and the mass of the electrons in the atom hardly plays a role:

Proton mass mp  =  1,672621898 × 10−24 g = 1.007276466889 u
Neutron mass mn  =  1,674927471 × 10−24 g = 1.008664915880 u
Electron mass me  =  9,109383560 × 10−28 g = 0.000548579909 u

When specifying the values, it is neglected that the atoms are also dependent on their energy, but this should not be discussed further here. Be in chemistry relative atomic masses used. The relative values ​​all relate to the value 1u.
 

isotope
Relative atomic mass of the isotopes
Carbon isotope C-12
12 u
Gold isotope Au-197
196,966569 u


196.966569 grams of gold contain as many atoms as 12 grams of the pure carbon isotope C-12. Many elements occurring in nature are mixed elements: they contain several different isotopes in their composition. While there is only one natural isotope in gold (Au-197), there are two natural isotopes in carbon, for example: C-12 and C-13. When specifying the relative atomic masses in the periodic table, the value refers to an average value of the atomic masses of the isotopes calculated according to the natural frequency. If only one isotope occurs in nature, then the relative atomic mass in the periodic table describes this isotope. If several natural isotopes occur in an element, then it is a mean value, calculated according to the frequency.


Element name in the PSE
Relative atomic mass of the elements
Carbon (2 isotopes)
12.01 u
Hydrogen (2 isotopes)
1.008 u
Oxygen (3 isotopes)
15,999 u
Gold (1 isotope Au-197)196,966569 u



Molar mass, molar mass

The molar mass can be derived from the basic quantities of mass and amount of substance M. be derived. The unit is KiIograms per mole (kg / mol) or grams per mole (g / mol). The relationship applies:

M.  =  m ÷ n = NA. × mM.

m Dimensions
n Amount of substance
NA. Avogadro's constant
mM. Particle mass

For an element that is not in molecular form, the numerical value is practically the same as the value given in the periodic table for the relative atomic mass. A gold atom has a relative atomic mass of 196.97u. So the molar mass for gold is:

M = 196.97 g / mol

With chemical compounds like H2O and also for elements that are in molecular form such as hydrogen H2 or oxygen O2, the value must be multiplied according to the number of atoms in the compound or the index numbers in the sum formula. The molar mass for water is:

M = 2 X 1.01 g / mol + 1 X 16.00 g / mol = 18.02 g / mol

 


1st exercise example, molar saltL.Prepare a solution with a certain concentration

Task: An aqueous sodium chloride solution is to be prepared which contains 1 mol of sodium chloride in one liter of solution. The concentration of the solution should be 1 mole per liter.

Method: Sodium chloride has the molecular formula NaCl. First, the relative atomic masses of the atoms involved from the periodic table are used.

M.NaCl = MN / A + MCl = 22.99 g / mol + 35.45 g / mol = 58.44 g / mol 

One liter of solution must therefore contain 58.44 grams of sodium chloride for the solution to contain the required concentration. The density of a saline solution does not correspond to the density of pure water, this must be taken into account. Therefore, the 58.44 grams of sodium chloride are first dissolved in half a liter of distilled water in a 1 liter volumetric flask. Only after the salt portion has been completely dissolved is the flask filled to the 1 liter mark.



2nd exercise example: Determine the amount of substance from a given portion of substance

taskssposition: What is the amount of substance approximately one gram of natural carbon?

Method: The equation for deriving the molar mass is first transformed. Then you can use the values ​​accordingly. So 1 gram of carbon is roughly one twelfth of a mole.
 
M = m ÷n | Reshaping
n = m ÷M.
n = 1g ÷12.01 mol / g ≈ 1/12 mol



3rd exercise example: diluting a concentrated solution

Task: Prepare an aqueous solution with a concentration of 1 mol per liter from 37% hydrochloric acid. Enter a recipe for this!

Method: First you have to calculate how many moles of hydrogen chloride HCl are dissolved in one liter of 37% hydrochloric acid. It should be noted that the density of 37% hydrochloric acid is 1.18 grams per cubic centimeter. The molar mass of hydrogen chloride is 36.46 grams per mole. Now you have to find out how many grams of hydrogen chloride are dissolved in one liter of 37% hydrochloric acid. The percentage is given in percent by mass: In one kilogram of 37% hydrochloric acid, 370 grams of hydrogen chloride would be dissolved. With the help of the equation for density you can first calculate how much 1 liter of 37% hydrochloric acid weighs:

D = m ÷V |
Reshaping
m = D
× V
m = 1.18 g / cm³X 1000 cm³ = 1180 g

So how much is 37% of 1180 grams? Multiplying the 1180 grams by 0.37 gives a value of 436.6 grams, which means that there are 436.6 grams of hydrogen chloride dissolved in one liter of 37% hydrochloric acid. Now you need the equation for the molar mass:

M = m ÷n | Reshaping
n = m ÷ M.
n = 436.6g ÷ 36.46 g / mol
≈ 12 mol

In one liter of 37%, fuming hydrochloric acid, about 12 moles of hydrogen chloride are dissolved. To prepare the solution 1 mole per liter, measure a twelfth of a liter in the fume cupboard (83 milliliters) of 37% hydrochloric acid in a measuring cylinder. This portion is dissolved in half a liter of distilled water in a volumetric flask and then made up to the 1 liter mark. Caution: "Never put water on the acid, otherwise the monster will happen!" Since it was rounded, the approach is not exactly exact, so that you only get an approximate solution with a concentration of 1 mol per liter. In this approach, a more precise calculation is made:

M = m ÷ n | Reshaping
m = M
× n
m = 36.46 g / mol
× 1 mol / l = 36.46 g

Inserting the values ​​in the rule of three:

1000 ml ÷ 436.6 g = x ml ÷ 36.46 g | Reshaping
x =
1000 ml x 36.46 g ÷ 436.6 g ≈ 83.5 ml
     


 
4. Üexercise example

The concentration of the dissolved sodium hydroxide of a sodium hydroxide solution (sample solution) is not known. The concentration should be determined with a neutralization by titration. During the titration, exactly 7 milliliters (7ml) of a 1 molar standard solution of hydrochloric acid are used until the neutralization point is displayed in the 20 milliliter (20ml) sample solution. What is the concentration of the caustic soda, neglecting the density?
 
Solution control
 


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